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Direction of the fastest rate of change of a function (Solution inclu)

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Find all the points at which the direction of the fastest rate of change of the function [itex]f(x,y)=x^{2}+y^{2}-2x-4y[/itex] is i+j

    2. Relevant equations

    3. The attempt at a solution
    The direction of the fastest change is [itex]\nabla f(x,y)=(2x-2)i+(2y-4)j[/itex], so we need to find all points (x,y) where [itex]\nabla f(x,y)[/itex] is parallel to i+j.

    [itex]\Longleftrightarrow(2x-2)i+(2y-4)j=k(i+j)[/itex] [itex]\Longleftrightarrow[/itex] [itex]k=2x-2[/itex] and [itex]k=2y-4[/itex]

    Ok here's where I start to get lost. How can k = two different things at the same time?

    The solution continues:
    Then [itex]2x-2=2y-4[/itex] [itex]\Longrightarrow y=x+1[/itex], so the direction of the fastest change is i+j at all points on the line y=x+1
  2. jcsd
  3. Apr 18, 2013 #2


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    Right. You can't have ##k## equal two different things. What you do have to do is figure out what relation ##y## and ##x## must satisfy so that you have only one value ##k## needed. That requires that ##2x-2=2y-4##.
  4. Apr 18, 2013 #3


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    The same way 4 can be equal to 2+ 2 and 3+ 1 "at the same time"= they are different ways of writing the same thing. That is why you can, as you say, write 2x- 2= 2y- 4.

  5. Apr 18, 2013 #4
    Ok, it just clicked.
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